Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 40}{x + 8} = \dfrac{x + 82}{x + 8}$
Explanation: Multiply both sides by $x + 8$ $ \dfrac{x^2 + 40}{x + 8} (x + 8) = \dfrac{x + 82}{x + 8} (x + 8)$ $ x^2 + 40 = x + 82$ Subtract $x + 82$ from both sides: $ x^2 + 40 - (x + 82) = x + 82 - (x + 82)$ $ x^2 + 40 - x - 82 = 0$ $ x^2 - 42 - x = 0$ Factor the expression: $ (x + 6)(x - 7) = 0$ Therefore $x = -6$ or $x = 7$ The original expression is defined at $x = -6$ and $x = 7$, so there are no extraneous solutions.